Question 1181767
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S12 = 2(3)^0+2(3)^1+2(3)^2+...+2(3)^(n-1)<br>
S12 = 2((3)^0+(3)^1+(3)^2+...+(3)^(n-1))<br>
Apply the formula for the sum of a finite geometric series:<br>
{{{S12 = 2((1-3^12)/(1-3)) = 3^12-1 = 531440}}}<br>
The second answer you show is correct.<br>
Your first answer is {{{2(3^12)}}}<br>
It is possible that for that answer you were using the correct formula but did the calculation incorrectly....<br>